# Example of normal distribution in an intervalNormal Distribution Example

### Problem Data

It is required to calculate the probability of P(485000<X<530000) (interval) with the following data:

Mean (μ)
500000
Standard Deviation (σ)
15000
Description
Example of normal distribution in an intervalA customer has an investment portfolio whose mean value is \$500,000 and whose standard deviation is \$15,000. You have asked him to calculate the probability that the value of his portfolio is between \$485,000 and \$530,000.

## Solution

Below are the detailed calculations and graphs to obtain the probability according to the data provided:

### Step 1:

We have the normal random variable X with mean μ and variance σ2; that is, X ~ N(μ,σ2), which will become the standard normal random variable, Z ~ N(0,1). To do so, we will apply the following transformation:

We have μ = 500000 and σ = 15000, then:

The exact result of the expression obtained is:

To check the result using the lower-tailed normal distribution (Z) table, we will make use of the following property:

In our problem it would be:

1. For P(Z<2)

The exact result of the expression obtained is:

To check the result we will use the left-tailed normal distribution (Z) table.

We will use our table to find the probability of P(Z<2.00) as follows:

• In the rows of the table we place the value of 2.0
• In the columns of the table we place the value of 0.00
• The intersection of the indicated row and column will give us the value of P(Z<2.00)=0.9772 (see highlighted value in the table below).

2. For P(Z<-1)

The exact result of the expression obtained is:

To check the result we will use the left-tailed normal distribution (Z) table.

Since our table does not show negative values, we will use the symmetry of the normal distribution to solve the problem:

We will use our table to find the probability of P(Z<1.00) as follows:

• In the rows of the table we place the value of 1.0
• In the columns of the table we place the value of 0.00
• The intersection of the indicated row and column will give us the value of P(Z<1.00)=0.8413 (see highlighted value in the table below).

Finally, we replace the value in the obtained expression:

Returning to the original expression, we replace the values obtained:

The probability that the random variable X lies in the interval [485000,530000] is 0.8186

Note: The results obtained by the calculator and the one obtained by the table may have small differences due to rounding issues.

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